Question

A car is moving on a circular level road of radius of curvature 300m. If the coefficient of friction is 0.3 and acceleration due to gravity is\[10\dfrac{m}{{{s^2}}}\]. The maximum speed the car can have is
A. 30 kmph
B. 81 kmph
C. 108 kmph
D. 162 kmph

Answer 1

Hint: In this question since coefficient friction is given so we will find the frictional force acting on the car and then we will equate it to the centripetal force and then we will find the maximum speed of the car on the circular road.

Complete step by step answer:
Let the mass of the car be ‘m’
Given
Radius of curvature of circular road\[r = 300m\]
Coefficient of friction is\[\mu = 0.3\]
Acceleration due to gravity \[g = 10\dfrac{m}{{{s^2}}}\]
We know the frictional force is given by the formula \[F = \mu mg - - (i)\]
Hence the maximum force applied on the car moving on a circular level road having coefficient of friction is\[\mu = 0.3\]will be equal to
\[
  {F_m} = \left( {0.3} \right)m\left( {10} \right) \\
   = 3mN \\
 \]
Now since the car is moving on the circular path so we will find the centripetal force on the car which is given by the formula \[{F_C} = \dfrac{{m{v^2}}}{r} - - (ii)\]
Hence by substituting the values in the equation (ii), we get
\[{F_C} = \dfrac{{m{v^2}}}{{300}}\]
Now we will equate frictional force with the centripetal force, so we can write
\[
  i = ii \\
\Rightarrow {F_m} = {F_C} \\
\Rightarrow \mu mg = \dfrac{{m{v^2}}}{r} \\
 \]
This can be further written as
\[
  3m = \dfrac{{m{v^2}}}{{300}} \\
 \Rightarrow 3 = \dfrac{{{v^2}}}{{300}} \\
\Rightarrow {v^2} = 900 \\
\Rightarrow v = 30\dfrac{m}{s} \\
 \]
Hence, the maximum speed the car can have is \[v = 30\dfrac{m}{s}\]
Also \[v = 30\dfrac{m}{s} = 30 \times \dfrac{{18}}{5} = 108\dfrac{{km}}{h}\]

So, the correct answer is “Option C”.

Note:
The frictional force supplies the centripetal force and is numerically equal to it. Students must note that the centripetal force is directly proportional to the velocity and uniform circular motion so larger will the centripetal force smaller will be the radius of curvature.